Harish Dorai wrote:
Problem #1:
Let us assume the height at the tree at the time planting as A which is equal to 4 Ft. Let us assume each year the tree grows by h feet.
So after
1st year height is A + h
2nd year height is A + 2h
.....
4th year A + 4h
6th year A + 6h
Given that after 6 years the height is 1/5 (which is 0.2) more than what it was in 4th year. So this can be converted into an equation as:
A + 6h = 1.2(A + 4h)
1.2h = 0.2A
h = A/6
= 4/6
= 2/3 feet.
Problem #2:
There is already a discussion in this forum that details the solution to the same problem. Please refer that.
Problem #3:
Let us assume for simplicity the speed (v-3) as X. So (v+3) = v-3 + 6 = X + 6
Time taken to travel upstream = 90/X
Time taken to travel downstream = 90/(X+6)
Given that upstream took 1/2 hour more.
So 90/X = 90/(X+6) + 0.5
If you simplify this, it becomes a Quadratic equation Square(X) + 30X - 1080 = 0
This is (X + 36) (X - 30) = 0. Which means X = -36 or X = 30
Since speed cannot be negative, the correct solution is X = 30
That is v-3 = 30
Which means v+3 = 36 mph.
Now time taken to travel downstream = 90/36 = 2 hours 18 minutes OR 2.3 hours.
Hope it helps.
How is 90/36 not 2.5 hrs?